Physical chemistry Flashcards

Question

3.1.2 The Ideal Gas Equation

Answer 1

The ideal gas equation is: pV = nRT, in this equation: * p = pressure (1atm = 101325Pa = 101325Nm-2) (Unit: Pa) * V = volume (1m3 = 1,000dm3 = 1,000,000cm3) (Unit: m3) * n = no. of moles (k = kilo = x1,000; M = mega = x1,000,000) (Unit: mol) * R = gas constant = 8.31JK-1 mol-1 * T = temperature (K = °C + 273; °C = K - 273) (Unit: K)

Answer 2

- Empirical formula: the simplest whole no. ratio of atoms of each element in a compound. - Molecular formula: the actual no. of atoms of each element in a compound. * Molecular formulae are simple whole no. multiples of empirical formulae.

Answer 3

1) Convert masses to moles: moles = mass/Ar 2) Divide each molar quantity by the smallest. 3) If one of the numbers are not close enough to round (not close enough to whole by 2 d.p.)multiply it by a factor accordingly.

Answer 4

Some chemical reactions (ones involving ionic compounds) are reactions between only some of the ions involved in a reaction. The balanced equation can be rewritten as an ionic equation leaving out the ions which do not take part in the reaction (Those non-partakers are called spectator ions)

Answer 5

Percentage yield: the percentage of the theoretical yield which is achieved in the reaction. * % yield = actual yield/theoretical yield x 100% - Theoretical yield: the amount of product that we expect to obtain based on our calculations. - Actual yield: the true amount of product obtained. --> Actual yield is less than the theoretical yield because: * loss by mechanical transfer (transferring from one container to another) * loss during a separating technique, e.g. filtration * side reactions occurring * reaction not being complete

Answer 6

Atom economy: a measure of how efficiently the atoms in the reactants are used in a chemical reaction. It can be calculated as a % using the following equation: % atom economy = Mr of desired product/sum of Mr of all reactants x 100%

Answer 7

concentration = moles of solute/volume of solution (dm3) This means: moles = [ ] x vol (dm3) (The units of [ ] are moldm-3 - also written as M - and gdm-3)

Answer 8

✴ Avogadro’s hypothesis: Equal volumes of any gas measured at the same temp and pressure contain the same number of moles of gas. Molar volume: the vol. of 1 mole of gas under the conditions of the reaction. moles = volume/molar volume (mol. = V/Vm) --> standard conditions 1) Standard Temperature and Pressure (STP), Vm = 22.4dm3 (273K (0°C), 1atm) 2) Room Temperature and Pressure (RTP), Vm = 24.0dm3 (298K (25°C), 1atm)

Answer 9

The change of state from solid to liquid. Temperature that this occurs = the melting point m.p.) * Energy is taken in when a substance melts to overcome bonds or intermolecular forces. The stronger the forces or bonds in a substance, the greater the energy required to melt the substance.

Answer 10

change of state from liquid to solid. Temperature that this occurs = the melting point also * Energy is released when a substance freezes as forces or bonds are formed. The stronger the forces or bonds formed on freezing, the more energy is released.

Answer 11

Change of state from liquid to gas. Temperature that this occurs = the boiling point (b.p.) * Energy is taken in when a substance boils. The stronger the bonds or intermolecular forces in a liquid substance, the greater the energy required to boil the substance.

Answer 12

change of state from gas to liquid. Temperature that this occurs = The boiling point (b.p.) * Energy is released when a substance condenses. The stronger the bonds or forces formed on condensing, the more energy is released.

Answer 13

Change of state from solid to gas on heating; change of state from gas to solid on cooling. * Substances which sublime for the A-Level = solid iodine and solid carbon dioxide (solid carbon dioxide is known as ‘dry ice’)

Answer 14

The electrostatic attraction between oppositely charged ions in an ionic lattice. These ions are formed by electrons being transferred from one atom to another.

Answer 15

Formulae of Compound Ions: Ammonium: NH4+ Carbonate: CO3 2- Hydroxide: OH- Nitrate: NO3- Sulfate: SO4 2-

Answer 16

1) Particles of the 3-D lattice are cations & anions. 2) Strong electrostatic attractions hold cations & anions together. Ionic compounds are hard, because the anions & cations are strongly attracted to each other therefore difficult to separate - but if force is applied and lattice shifts, same charges will repel

Answer 17

The giant ionic lattices are held together by a large number of strong electrostatic attractions between oppositely charged ions. Therefore, large amount of energy required to overcome the forces. That means ionic compounds tend to have high b.p.’s and m.p.s * Importantly, the smaller the ions (in terms of actual size) and the higher the charge on the ions, the stronger the ionic bond.

Answer 18

To conduct electricity, they must: 1. The substance must contain charged particles. 2. The charged particles must be free to move throughout the substance.

Answer 19

Ions are charged particles. BUT, they are fixed in positions in the lattice by strong ionic bond. This means ionic compounds do not conduct in solid state. * In molten state, or in an aqueous solution, the ions are free to move. So the solution containing the ions, e.g. salt water (NaCl in H2O), will conduct.

Answer 20

- Ionic compounds are polar molecules, this means they tend to dissolve in polar solvents, such as water - When moving water molecules hit an ionic lattice, they can knock ions off the outer layers - The water molecules then surround the ions, causing the substance to break up and dissolve

Answer 21

The electrostatic attraction between a bonding pair of electrons and the nuclei of the 2 atoms involved in the bond. * Atoms make 1 covalent bond for each electron that it needs to get a ‘full’ shell.

Answer 22

1) Particles of the 3-D lattice are covalent molecules. 2) The forces between the molecules are weak van der Waals forces. - for example I2 (Iodine)

Answer 23

- Relatively soft as they do not have strong intermolecular forces. - Relatively low because the van der Waals forces between the molecules are weak. This means it only takes a little energy required to overcome them.

Answer 24

- They’ll dissolve in non polar solvents, because they themselves are not polar. This means they will not dissolve in water. Your typical solvent will be some non-polar organic substances. - Covalent things do not conduct electricity; covalent things don’t contain any charged particles which are free to move.

Answer 25

1) The atoms are held together by strong covalent bonds in a giant network. - for example diamond and graphite

Answer 26

- Very hard due to many strong covalent bonds + rigid 3-D structure holding surface atoms in place. - You need very high temperatures to melt and boil giant covalent things. This is because ALL the strong covalent bonds holding the giant lattice have to be broken which needs a lot of energy. Compare this to simple covalent structures, where only VDW’s or Dipole Dipole IMF’s have to be broken and you quickly see why the M.P./B.P’s are so different

Answer 27

- These lot are insoluble. All the strong covalent bonds of the giant covalent structure need to be broken to free the atoms and let them move through the solvent. - There are no charged particles, since the lattice particles are atoms, so giant covalent crystals are insulators. Graphite is an exception to this rule because it has a ‘spare electron’. It doesn’t really break the rule as an electron IS a charged particle and in graphite's case, it can move!

Answer 28

It is the electrostatic attraction between metal cations (metal positive ions, cat means +) delocalised electrons in a lattice

Answer 29

1) Particles of the 3-D lattice are positive ions surrounded by a ‘sea’ of delocalised electrons. 2) The positive ions are held together by metallic bonding; metallic bonding literally is the electrostatic attraction between metal cations and the associated delocalised electrons.

Answer 30

- Metals are typically hard, due to strong electrostatic attraction between metal cations and delocalised electrons. - Metals are Insoluble, except in other liquid metals. This is, again, because of the strength of metallic bonds

Answer 31

Metals have relatively high M.P./B.P. because there are strong forces of attraction between cations and delocalised electrons * The greater the number of delocalised electrons per atom, the stronger the metallic bond so therefore the higher the m.p./ b.p. * The smaller the ions in the metallic lattice, the stronger the attraction between the delocalised electrons + the cations therefore the higher the m.p./b.p. * Transition metals have much higher m.p./b.p. than the main group metals (group 1 and 2) due to the large number of d sub-shell electrons. These electrons may become delocalised, creating an even stronger metallic bond. * This principle : smaller ion + more charge = stronger bond is the same as the ionic bonding principle

Answer 32

* Metals have delocalised electrons in their structures, these can move. An electric current = the movement of charged particles. * Heat is conducted when particles can move and are close enough together to pass the heat energy from one to another. The delocalised electrons enable heat energy to be passed through the metal because they’re nice and close. So metals are good conductor of heat.

Answer 33

- Because of the layered structure of the lattice, metallic layers can slide over each other without disrupting the bonding. This makes metals malleable - They’re also ductile (this means you can draw them out into a wire)

Answer 34

- An Alloy is simply a mixture of metals. - Alloy’s aren’t malleable because different metals are different in size - This makes the structure of an Alloy irregular, so it is much more difficult for layers to slide.

Answer 35

- The shapes of a covalent molecule or ion depends on the repulsion of the electrons around a central atom. - The electron pairs are charge clouds, areas where you have a big chance of finding an electron around an atom that repel each other.

Answer 36

There are 2 types of electron pairs: * Bonding pairs: pairs of electrons which are shared in covalent bonds. * Lone (non-bonding) pairs: pairs of electrons which are unshared.

Answer 37

✴ Pairs of electrons in the outer shell of atoms arrange themselves as far apart as possible to minimise repulsion. ✴ Lone pairs are held closer to the central atom so they have a greater repulsive effect on the other pairs of electrons. - This means lp–lp repulsion is greater than lp–bp repulsion, which is greater than bp–bp repulsion.

Answer 38

The molecule or ion will take up a shape which minimises electron pair repulsions. The shape of the molecule is determined from: * The total amount of electron pairs around a central atom. * The amount of (sets of) bonding pairs of electrons. * The amount of lone pairs of electrons.

Answer 39

- Linear ; BP=2, LP=0, BA= 180 - Trigonal planar ; BP=3, LP=0, BA=120 - Tetrahedral ; BP=4, LP=0, BA= 109.5 - Trigonal Bipyramidal ; BP=5, LP=0, BA= 90 + 120 - Octahedral ; BP=6, LP=0, BA= 90

Answer 40

- Pyramidal ; BP=3, LP=1, BA= 107 - Bent ; BP=2, LP=2, BA= 104.5 - Trigonal planar ; BP=3, LP=2, BA= 120 - Square planar ; BP=4, LP=2, BA= 90

Answer 41

These are forces that DON’T require the exchange of an electron. They exist because of bonds, they aren’t bonds themselves. They exist between molecule.

Answer 42

The measure of attraction by an atom to a pair of electrons in a covalent bond. * Electronegativity is measured on the Pauling Electronegativity Scale. A higher number on the scale means an element is better able to attract the bonding electrons in a covalent bonds; a higher number means it likes electrons more.

Answer 43

1) The distance of the bonding electrons from the attractive power of the nucleus (atomic radius). 2) The size of the nuclear charge (i.e. atomic no.). 3) The attractive power of the nucleus being shielded by inner electrons - shielding

Answer 44

INCREASES. This is because… * atomic radius goes down across a period, giving a progressively stronger attraction between the positive nucleus and the 2 electrons in the covalent bond. * nuclear charge goes up across a period. This will cause a greater attraction for the electrons in the covalent bond.

Answer 45

DECREASES. This is because: * atomic radius goes up down a group, giving a progressively weaker attraction between the positive nucleus and the electrons in the covalent bond. * shielding of the nuclear charge increases down a group as there are more electrons in inner energy levels.

Answer 46

The electronegativities of the 2 atoms in the covalent bond are the same. This means that the bonding pair of electrons are shared equally, i.e. electron distribution is symmetrical. E.g. H-H, O=O etc.

Answer 47

One atom is more electronegative than the other. The electrons in the covalent bond will therefore be closer to one atom than the other, i.e. electron distribution is unsymmetrical. This means one atom will be partially positive and the other partially negative. E.g. H-Cl, C-Cl, Be-Cl, C=O, etc.

Answer 48

It is a difference in electronegativity between 2 atoms caused by a shift in electron density in the bond.

Answer 49

* In a polar bond, the difference in electronegativity between the 2 atoms causes a dipole. * If charge is distributed unevenly over a whole molecule, then the molecule will have a permanent dipole. Molecules that have permanent dipoles = polar molecules. Whether or not a molecule is polar depends on whether it has any polar bonds + its overall shape.

Answer 50

- two polar C=O bonds exactly cancel each other out, so the molecule has no permanent dipole moment (O=C=O) - If the polar bonds are arranged so that they all point roughly in the same direction + they don’t cancel each other out, then charge will be arranged unevenly across the whole molecule = molecule has a permanent dipole + is polar.

Answer 51

These are attractive forces between all molecules and atoms caused by induced dipoles, which are temporary and constantly shifting.

Answer 52

* Electrons in charge clouds are always in rapid motion + at any one time they may be distributed more on one side of the molecule than the other. Another molecule approaching this side of the molecule will have its electrons repelled - this creates a temporary or induced dipole. * The 2 dipoles are then attracted to each other, and the 2nd dipole can cause yet another dipole in a 3rd atom. - larger the molecule ➜ higher the Mr ➜ greater the no. of electrons ➜ greater the induced dipoles ➜ greater the van der Waals forces between atoms

Answer 53

These are weak electrostatic forces of attraction between the δ+ end of the dipole on one molecule and the δ- end of the dipole on another neighbouring molecule.

Answer 54

* The permanent dipole-dipole forces are responsible for holding together the molecules in polar substance, but there are also VDW’s forces between these polar molecules as well.

Answer 55

If you put an electrostatically charged rod next to a jet of a polar liquid (e.g. water), the liquid will move towards the rod - this is because polar liquids contain molecules w/ permanent dipoles. * The more polar the liquid, the stronger the electrostatic attraction between rod + jet, so the greater the deflection will be. * By contrast, liquids made up of non-polar molecules (e.g. hexane) will not be affected at all when placed near a charged rod.

Answer 56

Hydrogen bonds occur between a δ+ H atom which is covalently bonded to O, N or F of one molecule + the lone pair of electrons of an O, N or F atom of another molecule. - Stronger than induced + permanent dipole-dipole forces.

Answer 57

* O, N + F are very electronegative, so they draw the bonding electrons away from the hydrogen atom. * This causes the bond to be very polarised, + because hydrogen has such a high charge density due to its small size, the hydrogen atoms form weak bonds w/ lps on F, N or O atoms of other molecules. * Only O-H and N-H groups are found in covalent molecules as H-F is only found in hydrogen fluoride.

Answer 58

The hydrogen bonds between water molecules explain many of its physical properties. - Higher boiling point + melting point than would be expected for group 6 hydrides

Answer 59

* The ↑ seen from H2S ➜ H2Se ➜ H2Te is caused by an ↑ in the van der Waals forces of attraction. As you descend group 6, each atom has more electrons so the induced dipoles increases. Hence van der Waals forces between molecules increase. * It would be expected based on the other hydrides that water should have a very low b.p. - this would be the case if the only intermolecular forces between water molecules were van der Waals forces of attraction + permanent dipole-dipole forces. * However, the hydrogen bonds between water molecules mean that it requires much more energy to break the bonds, hence giving it a higher b.p. + m.p.

Answer 60

As liquid water cools to form ice, the molecules make more hydrogen bonds and arrange themselves into a more open, regular 3-D lattice structure. - Since hydrogen bonds are relatively long, the average distance between water molecules is greater in ice than in liquid water - so ice is less dense than liquid water.

Answer 61

The heat energy change measured under constant pressure. (Unit: kJ mol-1)

Answer 62

These give out energy to their surroundings, so the temperature in the reaction goes up. The products of the reaction end up with less energy than the reactants. This means that the enthalpy change for the reaction, ∆H, will be negative.

Answer 63

These take in energy from their surroundings, so the temperature in the reaction falls. The products of the reaction end up with more energy than the reactants. This means that the enthalpy change for the reaction, ∆H, will be positive.

Answer 64

* Standard conditions are represented by the symbol Ɵ, (read ‘feye’) - Standard conditions are 100 kPa (about 1 atm) and 298K. * State symbols should always be included in chemical equations to represent enthalpy changes.

Answer 65

The enthalpy change when a reaction occurs in the molar quantities shown in the chemical equation, under standard conditions with all reactants and products in their standard states * E.g. CaO(s) + H2O(l) → Ca(OH)2(s)

Answer 66

The enthalpy change when 1 mole of a compound is formed from its constituent elements in their standard states under standard conditions. * E.g. 2C(s) + 3H2(g) + ½O2(g) → C2H5OH(l) - The enthalpy change of formation of an element is zero if the element is in its standard state. E.g. the enthalpy of formation of oxygen, O2(g) , is zero.

Answer 67

The enthalpy change when 1 mole of a substance is completely burned in oxygen under standard conditions w/ all reactants + products in their standard states. * E.g. CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)

Answer 68

They are experiments that measure the heat given out by reactions

Answer 69

Put the reactants (with a known conc or mass) in a polystyrene cup (to reduce heat loss or gain through the sides), that is in a beaker (for stability). Use a thermometer to measure the temperature of the mixture at regular intervals. Stir the solution to make sure it’s evenly heated.

Answer 70

Energy given out by a combustion reaction can be calculated by measuring the temperature change it causes as it burns. To find the enthalpy of combustion of a flammable liquid, you burn it in a calorimeter

Answer 71

* Heat is always lost to the surroundings, however well your calorimeter is insulated. * When you burn a fuel, some of the combustion that takes place may be incomplete * Flammable liquids are often quite volatile, so some of the fuel may be lost by evaporation.

Answer 72

1) Measure the temperature at regular intervals, beginning a couple of minutes before the start of the reaction so that you have recorded down temperatures before and after the reaction. 2) Plot the graph of your results. 3) Draw two lines of best fit: one going through the points before the reaction started, one going through the points after the reaction started. 4) Extrapolate both lines so that they both pass the time when the reaction started. 5) Distance between the 2 lines at the time the reaction started (before any heat was lost) is the accurate temp. change for the reaction.

Answer 73

1) Calculate the heat change, q, in a reaction. This is given by the equation q = mc∆T * Where m (unit: g) is the mass of the substance that has a temp. change ∆T (unit: K or °C) + a specific heat capacity c (unit: JK-1g-1). 2) Change the units of q from J to kJ, because standard enthalpies of combustion are always given in units of kJmol-1. 3) Calculate the no. of moles of fuel that caused this enthalpy change, from the mass that reacted. Use the equation moles = mass/Mr 4) Calculate the standard enthalpy change of combustion, ∆cHƟ (in kJ mol-1), using q (in kJ), + the no. of moles of fuel that burned. Use the equation: ∆cHƟ = q / no. of moles reacting

Answer 74

✴ The enthalpy change is independent of the reaction route taken

Answer 75

Hess’s Law says that the total enthalpy change for route 1 is the same as for route 2 - this is true so long as the starting and ending conditions are the same. Calculations using enthalpies of formation ∆H = [sum of ∆fHƟ products] - [sum of ∆fHƟ reactants] * ∆fHƟ for elements is zero, as the element is being formed from the element.

Answer 76

∆H = [sum of ∆cHƟ reactants] - [sum of ∆cHƟ products] * Remember that you can ignore ∆cHƟ for oxygen, because you can’t burn 1 mole of oxygen in oxygen. E.g. calculate ∆fHƟ of ethanol using e.g. enthalpy values: 2C(s) + 3H2(g) + ½O2(g) → C2H5OH(l) ∴ ∆H = ∆cHƟ [C] x 2 + ∆cHƟ [H2] x 3 - ∆cHƟ [C2H5OH] = (-394 x 2) + (-286 x 3) - (-1367) = -279 kJ mol-1

Answer 77

It is the energy required to break a bond.

Answer 78

It is the average energy required to break one mole of a covalent bond (in the gaseous state) averaged over a range of compounds containing the bond.

Answer 79

* Energy is required to break bonds ∴ bond breakage = endothermic ∴ ∆H = +ve. - Stronger bonds take more energy to break. * Energy is released to form bonds ∴ bond formation = exothermic ∴ ∆H = -ve. - Stronger bonds release more energy when they form. * The enthalpy change for a reaction is the overall effect of these two changes. If you need more energy to break bonds than is released when bonds are made, ∆H = +ve. If it’s less, ∆H = -ve.

Answer 80

* Generally the shorter the covalent bond, the stronger the bond. * Triple covalent bonds are generally shorter (+ hence stronger) than double covalent bonds which are shorter (+ hence stronger) than single covalent bonds.

Answer 81

∆H = sum of mean bond enthalpies of bonds broken - sum of mean bond enthalpies of bonds made

Answer 82

* The reactions are considered in the gas phase so that only the covalent bonds are involved in the calculations. In liquid or solid phase, intermolecular forces would also be involved.

Answer 83

* ∆H values calculated using this method will be less accurate than using enthalpy of formation or combustion data because the mean bond enthalpies are not exact. ∴ Any deviation of the answer to a calculation from the quoted value may be due to the mean bond enthalpy values used or the reactants not being in their gaseous states in the calculation.

Answer 84

Collision theory states that reactions only occur when collisions take place between particles with sufficient energy - this minimum energy is called the activation energy.

Answer 85

It is the minimum amount of energy reacting particles must have for a successful reaction to take place. - This means that most collisions do not lead to a reaction; only a small proportion of particles collide with energy that’s greater than the activation energy

Answer 86

It’s the change in concentration of a reactant or product in a given amount of time (concentration ÷ time)

Answer 87

- Concentration - Pressure - Temperature - Surface area of solid reactants - Presence of a catalyst

Answer 88

* If concentration of a reactant is increased, * More particles of that reactant are present in a given volume, leading to more successful collisions between reactant particles in a given period of time * This causes the rate of reaction to increase * Concentration UP = Rate UP

Answer 89

* If pressure on a gaseous reaction is increased, * The particles are forced closer together, * Leading to more successful collisions between reactant particles in a given period of time, * This causes the rate of reaction to increase * Pressure UP = Rate UP

Answer 90

* If temperature goes up, the particles gain more KE and move faster, * Leading to more frequent successful collisions between reactant particles in a given period of time, this causes the rate of reaction to increase * Temperature UP = Rate UP

Answer 91

* If the surface area of solid reactants increases, the exposed surface of the reactants increases too. * This increases the number of successful collisions in a given period of time * Which increases the rate of the reaction. * Surface area UP = Rate UP

Answer 92

* A catalyst increases the rate of reaction without being used up. * A catalyst works by providing an alternative reaction pathway of lower activation energy. * By lowering the activation energy, more collisions are successful in a given period of time, so the rate of reaction increases. - Catalysts are important in many industrial processes by speeding up the process + so reducing costs. - Catalyst added = Rate UP

Answer 93

It is a substance that increases the rate of a chemical reaction without being changed in chemical composition or amount.

Answer 94

a plot of the number of gaseous molecules against the energy they have at a fixed temperature. * A single plot on the graph shows the distribution of molecular energies at a constant temp.

Answer 95

The peak of the curve represents the most likely energy of any single molecule - most of the molecules are moving at a moderate speed so their energies are in the middle

Answer 96

It shows the mean (average) energy of all the molecules is a bit to the right of the peak

Answer 97

Some molecules have more than the activation energy. - These are the only particles/molecules that can react

Answer 98

The curve starts at (0,0) because no molecules have zero energy

Answer 99

- When temp. of the reactant molecules is increased, this will ↑ the energy of the gaseous reactant molecules, changing the shape of the Maxwell Boltzmann distribution. - A small ↑ in temp. can cause a large ↑ in the rate of reaction as there is a significant ↑ in the no. of molecules w/ enough energy to undergo a successful collision. - At higher temps., more molecules have at least the activation energy

Answer 100

* Lower temp. distributions are moved to the left + the peak is higher. * Higher temp. distributions are moved to the right + the peak is lower. * The graphs should only cross once. * The curves should always start at the origin + should end up being asymptotic to the energy axis at higher energy levels.

Answer 101

- Increasing the conc. means there are more molecules available to collide in a given volume . There will also be more molecules with energies above the Ea - Rate of reaction increases with an increase in concentration, but the effect is much less significant than is achieved when increasing the temperature. This is because only a few more reactant molecules with enough energy to undergo a successful collision.

Answer 102

* The curve retains the same basic shape. This means that the most probable energy of the molecules remains the same so the peak should be higher but at the same energy value on the horizontal axis. * As there are more reactant molecules at the same temp., the overall area under the curve increases. - This increases the number of reactant molecules which have enough energy to undergo a successful reaction, leading to an increases in the rate of reaction.

Answer 103

A catalyst increases the rate of reaction by providing an alternative reaction pathway with a lower activation energy. * The area above the activation energy with a catalyst is greater, indicating that there are more molecules with sufficient energy to undergo a successful collision. This increases the rate of reaction. - more molecules have at least the activation energy in the catalysed reaction

Answer 104

* Adding a catalyst to a reaction mixture lowers the activation energy *A catalyst does not affect the shape of the distribution as long as the temperature and total number of molecules are not changed. - the line of the activation energy point will be moved more to the left

Answer 105

* The forward and reverse (backwards) reactions proceed at equal rates * the concentrations of reactants and products remain constant. A dynamic equilibrium can only happen in a closed system which is at constant temperature.

Answer 106

✴ If a factor is changed which affects a system in equilibrium, the position of equilibrium will shift in the direction opposing that change.

Answer 107

* An increase in pressure at constant temperature shifts the position of equilibrium to the side with fewer moles of gas. This is opposing the change by decreasing the pressure of the system. * Logically, a decrease in pressure at constant temperature shifts the position of equilibrium to the side with more moles of gas. This opposes the change by increasing the pressure.

Answer 108

* an increase in temperature would shift the position of equilibrium in the direction of the reverse endothermic reaction. This moves the position of equilibrium from right to left in order to oppose change, because the reverse reaction absorbs heat. (it’s endothermic!) * a decrease in temperature would shift the position of equilibrium in the direction of the forward exothermic reaction. The position of equilibrium would move from left to right, opposing the change and increasing temperature.

Answer 109

* an increase in temperature would shift the position of equilibrium in the direction of the forward endothermic reaction and the position of equilibrium would move from left to right in order to oppose change. Decreasing the temperature by absorbing heat. * a decrease in temperature would shift the position of equilibrium in the direction of the reverse exothermic reaction and the position of equilibrium would move from right to left. This opposes the change and increases the temperature by giving out heat.

Answer 110

* An increase in concentration of the reactants would shift the position of equilibrium from left to right in order to oppose change by increasing the yield of products and, therefore, decreasing the concentration of reactants. * An increase in concentration of the products would shift the position of equilibrium from right to left in order to oppose change by increasing the yield of reactants and therefore decreasing concentration of products.

Answer 111

* A catalyst increase the rate of the forward and reverse reactions equally therefore allowing equilibrium to be reached faster. * However, it has no effect on the position of equilibrium.

Answer 112

Ethanol can be produced by the hydration of ethene with steam, which is a reversible reaction: C2H4(g) + H2O(g) ⇌ C2H5OH(g) ∆H = -46kJmol-1

Answer 113

1) Concentrated phosphoric acid catalyst. 2) High pressure - (60-70atm.) 3) Temperature of 300°C 4) Excess ethene/steam can be used 5) Ethanol can be condensed out of the reaction mixture

Answer 114

* Allows equilibrium to be attained more rapidly as catalyst increasing the rate of reaction (hence also allowing lower temperature to be used, meaning lower energy costs). * No effect on position of equilibrium, therefore no effect on yield of product

Answer 115

* Increases equilibrium yield of ethanol (as 2 moles on the left and 1 mole on the right, therefore position of equilibrium shifts to the right) and increased rate of reaction (think collision theory) * However, too high a pressure is expensive (increased electrical pumping costs and requires expensive equipment to withstand the pressure.)

Answer 116

* Forward reaction = exothermic, so increased temperature will shift position of equilibrium in the direction of the reverse endothermic reaction leading to a lower yield. SO lower temperature means higher yield. * However, a lower temperature gives a slower rate of reaction, again, think collision theory * So in conclusion: 300°C is a compromise temperature; it’s a compromise between high rate and high yield.

Answer 117

* Excess ethene/steam will shift the position of equilibrium to the right to remove the excess reactant so increase yield of ethanol. * Any unreacted ethene is recycled back into the reaction mixture to increases overall yield.

Answer 118

* If a product is removed from an equilibrium reaction, the position of equilibrium will shift to the right to replace it, therefore increasing the overall yield.

Answer 119

* A homogeneous reaction is one in which all the reactants and products are in the same physical states. * A heterogeneous reaction (hetro meaning different, geneous meaning state) is one in which all the reactants and products are NOT in the same physical states.

Answer 120

* A SQUARE BRACKET [] AROUND A CHEMICAL MEANS CONCENTRATION OF THAT CHEMICAL. * With that, at dynamic equilibrium, the ratio: [reactants]:[products] is constant. The ratio’s value is known as Kc. (Constant of concentration) It’s always the same for a particular reaction under the same conditions.

Answer 121

Kc = [products] [reactants]

Answer 122

* If Kc>1, [products]>[reactants] and we say that the equilibrium lies to the RH side. * If Kc<1, [products]<[reactants] and we say that the equilibrium lies to the LH side.

Answer 123

It can be calculated for reactions in solution OR homogeneous gaseous reactions, as the concentration of a solution or gas can be calculated as the number of moles in a certain volume (in dm3).

Answer 124

For the general reaction: aA + bB ⇌ cC + dD Kc = [C]c[D]d [A]a[B]b

Answer 125

The units are in terms of conc. in mol dm3 but the overall power depends on the balancing no.s in the eq. for the reaction. SO, units of Kc = (mol dm-3)(c+d) (mol dm-3)(a+b) Therefore, if Kc has no units: there are equal numbers of moles on both sides of the reaction equation.

Answer 126

Kc will only change when temperature changes - other factors have no effect on Kc.

Answer 127

Oxidation state: the hypothetical charge an atom has, assuming that the bonding is completely ionic. * It is a numerical value of the ‘degree of oxidation or reduction’ of an atom.

Answer 128

* Uncombined elements have an oxidation state of 0. (An atom of lithium, for example, has an oxidation state of 0) - Elements bonded to identical atoms also have an oxidation state of 0. (for example both oxygen atoms in the Oxygen-Oxygen / O2 molecule) * The oxidation state of a simple monatomic ion is the same as its charge. (For example, an Li+ ion has an oxidation state of 1)

Answer 129

* The total sum of the oxidation states of all the elements in a compound must add up to zero if it’s not an ion. (for example: NaCl, Sodium is Na+ and chlorine (chloride) is Cl- the oxidation state of sodium is +1 and chlorine (chloride) is -1. +1 + -1 = 0) * The total sum of the oxidation states of all the elements in a molecular ion must add up to the charge on the ion. (for example: SO42-. Oxygen has a default 2- oxidation state, so 4 x -2 = -8 and, therefore, sulphur has an oxidation state of +6)

Answer 130

* Following on from that example, Oxygen has an oxidation state of -2 in all compounds except in peroxides, where it has an oxidation state of -1 and in OF2 where it has an oxidation state of +2. * Hydrogen has an oxidation state of +1 in all compounds except in hydrides where it has an oxidation state of -1. * Group 1 elements have an oxidation state of +1 in all compounds. * Group 2 elements have an oxidation state of +2 in all compounds. * The oxidation states of transition elements and p block elements vary. * In a simple binary compound (2 elements), the more electronegative element has the negative oxidation state.

Answer 131

E.g. find the oxidation state of Zn in Zn(OH)2. * Zn(OH)2 has an overall charge of 0 therefore overall oxidation state = 0 * Oxidation state of O = -2 and oxidation state of H = +1 * Hence oxidation state of Zn in Zn(OH)2 = 0 - 2 x (-2 + 1) = +2

Answer 132

* The charge of an ion of a d block element is given using Roman numerals. They decided to be annoying and can have multiple oxidation states (e.g. copper(II) = +2, in the same breath copper(III) = +3).

Answer 133

An increase in oxidation state is oxidation; a decrease in oxidation state is reduction. * Oxidation: the process of electron loss. * Reduction: the process of electron gain. * OIL RIG: Oxidation is loss, Reduction is gain (of electrons) A redox reaction is one in which an oxidation + reduction occur simultaneously in the same reaction.

Answer 134

In the reaction: 6HI + H2SO4 → 3I2 + S + 4H2O * Iodine (I) is oxidised, it shows an increase in oxidation state: from -1 (in HI) to 0 (in I2). * S is reduced as it shows a decrease in oxidation state: from +6 (in H2SO4) to 0 (in S). * This means (UNBELIEVABLY!) that this is a redox reaction; both oxidation and reduction occur together in the same reaction.

Answer 135

In the reaction: Cl2 + H2O → HCl + HOCl * Cl is oxidised as it shows an increase in oxidation state from 0 (in Cl2) to +1 (in HOCl). * Cl is also reduced as it shows a decrease in oxidation state from 0 (in Cl2) to -1 (in HCl). * This is a redox reaction, but a special type called a DISPROPORTIONATION REACTION, as chlorine is oxidised and reduced in the same reaction; for it to be a disproportionation reaction, the same species must be oxidised and reduced.

Answer 136

* Oxidising agent: an electron acceptor. They cause another substance to lose electrons. They oxidises another thing, but are reduced themselves * Reducing agent: an electron donor. - They cause another substance to gain electrons. They reduce another thing, but are oxidised themselves

Answer 137

Half-equations show oxidation or reduction. The electrons are shown in a half-equation so that the charges balance. E.G. * Oxidation of iron: Fe → Fe3+ + 3e- * Reduction of chlorine: Cl2 + 2e- → 2Cl

Answer 138

Half-equations for oxidising and reducing agents can be combined to make full equations. - However, make sure before combining the two that both half-equations have the same number of electrons in them (when they do, combine them and cancel out on each side).

Answer 139

Half-equations can be written to identify the oxidation and reduction processes in redox reactions. - When doing so, make sure the atoms and charges balance.

Answer 140

Sometimes half-equations need to be written for a more complicated reaction where the oxidising or reducing agent contains oxygen or hydrogen. If so, H2O and H+ ions may need to be added to balance the equation.

Answer 141

The rate (concentration change ÷ time change) of a chemical reaction is related to the concentration of reactants by a rate equation of the form: Rate = k [A]m [B]n

Answer 142

The power to which the concentration of a reactant is raised in the rate equation. ➜ The orders m and n are restricted to the values 0, 1, and 2 for the exam.

Answer 143

* Where m and n are the orders of reaction with respect to reactants A and B. K is known as the rate constant, it’s the same for a given temperature.

Answer 144

The proportionality constant which links the rate of reaction to the concentrations in the rate equation

Answer 145

rate = k [B]2 [D] This means: * the order of reaction with respect to B is 2. * the order of reaction with respect to D is 1. * the order of reaction with respect to C is 0.

Answer 146

* This is because [C] does not appear in the rate equation; if you do ALevel maths you’ll know anything ‘to the power of 0’ is — having [C]0 is the same as multiplying everything by 1, so we leave it out.

Answer 147

The overall order is all the ‘orders’ added up: 2 + 1 + 0 = 3

Answer 148

- Doubling the conc. of B would quadruple the rate as (2)2 = 4. - Doubling the conc. of C would have no effect on the rate as (2)0 = 1. - Doubling the conc. of D would double the rate as (2)1 = 2. - Doubling the conc. of B, C, and D would increase the rate by a factor of (2)3 = 8.

Answer 149

* The units of rate are mol dm-3 s-1 | I.E. concentration ÷ time * the units of conc. are mol dm-3 * The units of the rate constant K depend on equation

Answer 150

➜ the overall order = 3 ➜ If we replace everything with units: ➜ mol dm-3 s-1 = k (mol dm-3)3 ➜ expand those brackets: ➜ mol dm-3 s-1 = k * (mol dm-3 * mol dm-3 * mol dm-3) ➜ Simplify ➜ mol dm-3 s-1 = k * mol3 dm-9 ➜ Then rearrange that equation so we have “k = something” ➜ k = mol dm-3 s-1 ÷ mol3 dm-9 ➜ When you divide by an index (power) you take them away So, ➜ K = mol dm-3 s-1 ÷ mol3 dm-9 BECOMES: K = = mol-2 dm6 s-1 ➜ Final units for K ➜ mol-2 dm6 s-1

Answer 151

* Therefore: when overall order = 1, units of rate constant = s-1 * And when overall order = 2, units of rate constant = mol-1 dm3 s-1

Answer 152

The rate constant is dependent on temperature; if a constant temperature is not been maintained, the rate will vary based on changes in concentrations of reactants as well as temperature * As temperature increase, the rate constant increases too.

Answer 153

The Arrhenius equation links the rate constant with activation energy and temperature. It’s written below: k = Ae–Ea/RT

Answer 154

* The mole fraction of one of the gasses, A, in a mixture of gasses is written as: xA. * It’s calculated by dividing the amount, in moles, of A by the total amount of moles of gas in the mixture.

Answer 155

The partial pressure of one of the gasses in a mixture of gases is the contribution which that gas makes to the total pressure of the gas mixture. * The partial pressure of A is denoted p(A) and its units are Pa (or kPa = 103 Pa)

Answer 156

The equilibrium constant Kp is deduced from the equation for a reversible reaction occurring in the gas phase. Kp is calculated from partial pressures for a system at constant temperature (in a very similar way to Kc)

Answer 157

Kp = p(C)^c p(D)^d/p(A)^a p(B)^b Therefore, units of Kc = (Pa)^(c+d)/(Pa)^(a+b)

Answer 158

* Kp is affected by changes in temp. only.

Answer 159

* Whilst a catalyst can affect the rate of attainment of an equilibrium, it does not affect the value of the equilibrium constant.

Physical chemistry Flashcards

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